Examples

VANDERBILT UNIVERSITY MATH 196 — DIFFERENTIAL EQUATIONS WITH LINEAR ALGEBRA EXAMPLES OF SECTION 5.5. Question 1. Write the form of the particular solution for the equations below (you do not have to find the values of the constants). (a) y 00 + 9y = 2 cos(3x) + 3 sin(3x). (b) y 000 + y 0 = 2 − sin x. (c) y (4) − 2y 00 + y = xex . (d) y 00 + 9y = 2x2 e3x + 5. Question 2. In the questions below, two linearly independent solutions of the associated homogeneous equation are given. Find the particular solution yp . (a) x2 y 00 − 4xy 0 + 6y = x3 , y1 = x2 , y2 = x3 . (a) (x2 − 1)y 00 − 2xy 0 + 2y = x2 − 1, y1 = x, y2 = 1 + x2 . Question 3. Show that the formula given in class for yp in fact produces a particular solution. SOLUTIONS. 1a. The homogeneous equation is y 00 + 9y = 0, with characteristic equation λ2 + 9 = 0, whose roots are ±3i. Hence y1 = cos(3x), y2 = sin(3x), are solutions of the homogeneous equation. Given the form of f (x), we look for yp = xs A cos(3x) + B sin(3x) . Since cos(3x) and sin(3x) are solutions of the homogeneous equation, we need s = 1, so yp = x A cos(3x) + B sin(3x) . 1b. The homogeneous equation is y 000 + y 0 = 0, with characteristic equation λ3 + λ = 0, whose roots are λ = 0 and λ = ±i. Hence y1 = 1, y2 = cos(x), y3 = sin(x), are solutions of the homogeneous equation. Given the form of f (x), we look for yp = xs A + xr B cos(x) + C sin(x) , 1

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MATH 196 - EXAMPLES SECTION 5.5

where A corresponds to 2 and B cos(x) + C sin(x) to sin(x). If s = 0 then A =constant repeats the solution y1 , while if r = 0 then cos(x) and sin(x) repeat y2 and y3 . Hence we set s = 1 and r = 1: yp = Ax + x B cos(3x) + C sin(x) . 1c. The homogeneous equation is y (4) − 2y 00 + y = 0, with characteristic equation λ4 − 2λ2 + 1 = 0, whose roots are ±1, each repeated twice. Hence y1 = e−x , y2 = xe−x , y3 = ex , y4 = xex , are solutions of the homogeneous equation. Given the form of f (x), we look for yp = xs Ax + B ex . Since ex and xex are solutions of the homogeneous equation, we need s = 2, so yp = x2 Ax + B ex . 1d. The homogeneous equation is y 00 + 9y = 0, with characteristic equation λ2 + 9 = 0, whose roots are ±3i. Hence y1 = cos(3x), y2 = sin(3x), are solutions of the homogeneous equation. Given the form of f (x), we look for yp = xs A + xr Bx2 + Cx + D e3x . Since there is no repetition with the solutions of the homogeneous equation, r = s = 0 and yp = A + Bx2 + Cx + D e3x . For question 2, recall that the formula for yp is Z Z y2 (x)f (x) y2 (x)f (x) dx + y2 (x) dx. yp (x) = −y1 (x) W (x) W (x) But it is very important to also remember that this formula is for an equation written with the coefficient of y 00 being one. If this is not the case we have to first divide the equation by the coefficient of y 00 . 2a. Rewrite the equation as 4 6 y 00 − y 0 + 2 y = x, x x so that f (x) = x. Compute W (x) = y1 y20 − y2 y10 = x2 (3x2 ) − x3 (2x) = x4 . Plugging all quantities into the formula for yp and performing the integrals we find yp = x3 (ln x − 1).


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2b. Rewrite the equation as y 00 −

2x 0 2 y + 2 y=1 −1 x −1

x2

so that f (x) = 1. Compute W (x) = y1 y20 − y2 y10 = x(2x) − (1 + x2 )1 = x2 − 1. Plugging all quantities into the formula for yp and performing the integrals we find 1 + x 1 2 + (1 + x2 ) ln |1 − x2 |. yp = −x + x ln 1 − x 2 3. We want to show that Z yp (x) = −y1 (x)

y2 (x)f (x) dx + y2 (x) W (x)

Z

y2 (x)f (x) dx. W (x)

solves y 00 + py 0 + qy = f, where p, q and f are not necessarily constants and y1 and y2 are solutions of the associated homogeneous equation, i.e., they solve y 00 + py 0 + qy = 0. We have to plug yp into the equation. Since we shall differentiate yp , it is useful to remember the Fundamental Theorem of Calculus, which gives Z y (x)f (x) 0 y (x)f (x) 2 2 dx = , W (x) W (x) and Z y (x)f (x) 0 y (x)f (x) 2 2 dx = . W (x) W (x) Using these formulas and the product rule we find Z Z y2 f y2 f y1 f y1 f 0 0 0 yp = −y1 − y1 + y2 + y2 , W W W W R 2f R (x)f (x) where we write yW instead of y2W (x) dx in order to simplify the notation (analogously for the other integral). Taking another derivative Z Z y2 f y2 f 0 y1 f y1 f 0 00 00 0 y2 f 00 0 y1 f yp = −y1 − 2y1 − y1 + y2 + 2y2 + y2 . W W W W W W Using yp , yp0 and yp00 into the equation we find yp00

+

pyp0

Z

Z y2 f y2 f 00 0 + qyp = − + + qy1 + y2 + py2 + qy2 W W 0 y2 f y1 f y2 f y1 f 0 0 0 − py1 + 2y1 + py2 + 2y2 − y1 + y2 . W W W W y100

py10

Since by hypothesis y1 and y2 are solutions of the homogeneous equation, y100 + py10 + qy1 = 0,

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and y200 + py20 + qy2 = 0, so y1 f y2 f y2 f 0 y1 f 0 0 0 + + qyp = − py1 + 2y1 + py2 + 2y2 − y1 + y2 W W W W By the product rule 0 y2 f 0 1 0 f 0 1 = y2 , + y2 f + y2 f W W W W and 0 y1 f 0 1 0 f 0 1 = y1 + y1 f + y1 f . W W W W Therefore y1 f y2 f py1 + 2y10 + py2 + 2y20 yp00 + pyp0 + qyp = − W W 0 f 1 1 − y1 y20 − y1 y2 f 0 − y1 y2 f W W W 0 1 0 1 0 f + y2 y1 f + y2 y1 f . + y2 y1 W W W Notice that the last two terms of the second line cancel with the last two terms of the third line. We are left with y1 f y2 f f f yp00 + pyp0 + qyp = − py1 + 2y10 + py2 + 2y20 − y1 y20 + y2 y10 W W W W y2 f y2 f 0 y1 f y1 f 0 f f = −p y1 − 2 y1 + p y2 + 2 y2 − y1 y20 + y2 y10 . W W W W W W The first and third terms on the last line cancel out, and then y1 f 0 f f y2 f 0 y1 + 2 y2 − y1 y20 + y2 y10 yp00 + pyp0 + qyp = −2 W W W W f f f f = −2 y2 y10 + 2 y1 y20 − y1 y20 + y2 y10 W W W W f f f 0 0 0 0 = y1 y2 − y2 y1 = y1 y2 − y2 y1 W W W = f, yp00

pyp0

where in the last step we used that W = y1 y20 − y2 y10 .